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Question:

A photon of wavelength 4 × 10–7 m strikes on metal surface, the work function of the metal being 2.13 eV. Calculate

 

(i) the energy of the photon (eV),

 

(ii) the kinetic energy of the emission, and

 

(iii) the velocity of the photoelectron (1 eV= 1.6020 × 10–19 J).

Answer:

(1)We know λ = 4 × 10–7 m(given)

C = 3 x 108

From the equation E= hv or hc/ λ

Where, h = Planck’s constant = 6.626 × 10–34 Js

c = velocity of light in vacuum = 3 × 108 m/s

λ = wavelength of photon = 4 × 10–7 m

Substituting the values in the given expression of E:

Hence, the energy of the photon is 4.97 × 10–19 J.

 

(ii) The kinetic energy of emission Ek is given by

= (3.1020 – 2.13) eV

= 0.9720 eV

Hence, the kinetic energy of emission is 0.97 eV.

 

(iii) The velocity of a photoelectron (ν) can be calculated by the expression,

 

 

Where, (hv-hv0) is the kinetic energy of emission in Joules and ‘m’ is the mass of the photoelectron. Substituting the values in the given expression of v:

v = 5.84 × 105 ms–1

Hence, the velocity of the photoelectron is 5.84 × 105 ms–1.

 


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